Algorithms – How to Use the 0-1 Knapsack Problem with Multiple Bags

It might be some mistake in your code. I wrote the DP program as mentioned by Naryshkin. For the target span 13, it reports 6+7, and for 15, it reports 2+6+7.

# weight: cable length
# total weight: target span
# value: 1 for each cable
# want minimum number of cables, i.e. minimum total value

def knapsack_01_exact_min(weights, values, W):
    # 0-1 knapsack, exact total weight W, minimizing total value
    n = len(weights)
    values = [0] + values
    weights = [0] + weights
    K = [[0 for i in range(W+1)] for j in range(n+1)]
    choice = [[0 for i in range(W+1)] for j in range(n+1)]
    for i in range(1, n+1):
        for w in range(1, W+1):
            K[i][w] = K[i-1][w]
            choice[i][w] = '|'
            if w >= weights[i]:
                t = K[i-1][w-weights[i]]
                if (w==weights[i] or t) and (K[i][w]==0 or t+values[i] < K[i][w]):
                    choice[i][w] = '\\'
                    K[i][w] = t+values[i]
    return K[n][W], choice

def print_choice(choice, weights):
    i = len(choice)-1
    j = len(choice[0])-1
    weights = [0] + weights
    while i > 0 and j > 0:
        if choice[i][j]=='\\':
            print weights[i],
            j -= weights[i]
        i -= 1
    print

lens = [10, 7, 6] + 5*[3] + 6*[2] + 7*[1]
values = (3+5+6+7)*[1]
span = 13
v, choice = knapsack_01_exact_min(lens, values, span)
print "need %d cables to span %d:" % (v,span),
print_choice(choice, lens)

span = 15
v, choice = knapsack_01_exact_min(lens, values, span)
print "need %d cables to span %d:" % (v,span),
print_choice(choice, lens)

If you adjust the order of the input lengths, it can give other optimal solutions. For example, lens = 5*[3] + 6*[2] + 7*[1] + [10, 7, 6] will give 15=10+2+3.

Bin packing is very computationally difficult. Think of half of the problem: you want to pack product in shipping boxes with no wastage in the box. An optimal solution for that would require going through all possible subsets and all possible 3d arrangements of the product that needs to ship in one truck. I'll give you the optimal solution for that because I have a friend who does six impossible things before breakfast.

Now you just have to get all the boxes on the truck with no wastage. My friend does his second impossible thing and gives you the solution. Unfortunately, with the box sizes you selected above, there is empty space in the truck which could be reduced if you'd chosen different (either larger or smaller) boxes in the first task. If you change the size of one box, at best you'll have to re-pack the truck; at worst, you may have to repack all of the boxes which is just as hard as the problem we started with. And, as with the first stage, you'd have to try all possible 3d arrangements.

I found Skiena's The Algorithm Design Manual to be helpful for thinking about what class of algorithms suit which sorts of problems, but I mostly learned that good solutions for even mundane problems blow up in you face with computational difficulty. Most of what you are needing fit into the class of bin-packing problems and that article is a good jumping off point. It is worth noting that some of the best algorithms for this are commercial products because this task pops up everywhere in logistics (what's the smallest number of train cars can I get my goods into? and such). There is considerable money to be made if the right heuristics can save a manufacturer 100 train cars a month.

Unfortunately, the literature on optimizing heuristics isn't nearly as large as for algorithms. If you try to go it alone, I guarantee that you'll be dreaming about moving rectangular prisms around by your second month. I had a cutting-stock problem that if I had to do again I'd probably farm out to the experts (or their propriety software).

Thanks to @JTrana for the fine expansion of my comment.